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As usual the puzzle was contributed by Mr Thomas Masiwa. A prize of $1000 (sponsored by Professor A.G.R.Stewart) was offered for the first correct solution opened after 31st August 2003. But nobody has won the prize.
Recall that we considered a strictly increasing sequence of natural numbers
{xn} such that:
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Across
1. limn® ¥ [(fn+1)/(fn)]=[(1+Ö5)/2] » 1,618 » 1.62
Note that limn® ¥([(1-Ö5)/2])n=0, since -1 < [(1-Ö5)/2] < 0.
3. First, find positive integral solutions to 120=8x2+5x1, which turn out to be x1=8 and x2=10. Next, find n=8 by solving the double quadratic equation, and then work out x8=13·10+8·8=194.
Down
1. We must apologise, for there was an error in the condition to be
satisfied by n. But that should not have stopped you, for after getting
the values x1=8 and x2=9 from the given x7=112, the only possible
value of n which will yield a number less than 200 is 8, and x8=181.
2. There are only two nontrivial factors of 91, namely 7 and 13, and only two perfect numbers less than 100, namely 6 and 28. Putting n=7 gives us x7=8·28 +5·6=274.
This sequence is the famous Fibonacci sequence. For the proof by induction, see similar problems in Issue 7.2, pages 17 and 26. the method of generationg functions.
